Suppose $(\Omega, \mathcal{\Sigma})$ is a measurable space equipped with two probability measures $P_1$ and $P_2$. $\mathcal{F}$ is a field in $\Omega$ and $\sigma(\mathcal{F})=\mathcal{\Sigma}$. (Added: A field of sets is defined to be closed under finite union and complement and contain $\Omega$.)

Thank you very much! 1) Is $\lim_{\delta \rightarrow 0} \quad \sup_{B \in \mathcal{F}, P_2(B) < \delta} P_1(B) = 0$ a limit in some sense? In your second paragraph, it seems to me it is like a limit of function. 2) Generally speaking, is this way of constructing $\mathcal{M}$ worth memorizing and has it been used in proving some theorems?
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steveOMay 2 '11 at 14:47

@steveO: 1) Sure, it's the limit as $x \to 0$ of the function $f(x) = \sup_{B \in \mathcal{F}, P_2(B) < x} P_1(B)$ :-) 2) I don't know about this particular $\mathcal{M}$, but monotone class theorem proofs tend to have this form: to show all sets in a $\sigma$-field $\Sigma$ have some property $Q$, show that the collection $\mathcal{M}$ of all sets with property $Q$ form a monotone class, then show there is a field $\mathcal{F}$ which generates $\Sigma$ and such that all sets in $\mathcal{F}$ have property $Q$.
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Nate EldredgeMay 2 '11 at 16:04

Thanks! How do you go from "for all $B \in \mathcal{F}$ with $P_2(B) < \delta$, we have $P_1(B) \le \epsilon$", to "for all $B \in \mathcal{M}$ with $P_2(B) < \delta$, we have $P_1(B) \le \epsilon$"?
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steveOMay 2 '11 at 16:36

@steveO: I'm not sure what you're asking. That comes from the way I defined $\mathcal{M}$. Indeed, I specifically chose the definition of $\mathcal{M}$ so that it would be true.
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Nate EldredgeMay 2 '11 at 16:56